Pressure Problem

An analysis of air containment in the Mars house

By IonMars 5-15-2014

The Mars house was designed to withstand compressive forces and the size of blocks was chosen for easy handling.  A question may arise concerning the ability of the house to withstand the stress of air pressure within the house when the outside air pressure is near zero.  We want the colonists to live and work in shirtsleeve conditions as much as possible and avoid the confinement of a spacesuit as much as possible.  The most comfortable pressure for a human is at Earth sea level, defined as 1 atmosphere (1atm= 14.7psi= 101.4kPa). (1) This represents a considerable pressure when there is no counter-pressure from the outdoor air.  It is comparable to an automobile tire that must contain pressure when inflated to 29.4 psi with a countervailing outside air pressure of 14.7 psi (29.4 psi - 14.7 psi = 14.7 psi).  We want the Mars house to contain air pressure at least as well as a tire.  This is not a trivial problem considering that a tire is designed with many cross woven threads and steel wires to withstand these pressures.

What stresses will be expressed at the interior ceiling of the Mats house when the inside pressure is at 14.7 psi?  The problem can be set up as the analysis of stress in a cylinder with enclosed ends.   Note that the shape of the top half of the house with an arched ceiling is the shape of 1/2 a cylinder. This is the critical part of the structure because the stone blocks comprising the ceiling are only 30 cm thick.  The sidewalls are also 30 cm thick but backed by a horizontal mass of compressed regolith greater than 2.5 meters.  The floor is thin, but supported underneath by the mass of the planet.  Thus the inside radius of the cylinder to be analyzed is 200 cm and the outside radius is 230 cm.  The inside length is 2030 cm.  Using Lame’s formulation for thick walled cylinders, the radial (hoop) force is calculated as 107 psi.  (See the calculation on page 3.)  The tangential stress and the axial stress can also be calculated but the radial stress is always the largest (2) and it will be considered the controlling factor for the safety of the house.  Using this formulation we find the radial stress to be 7.51 kg/cm2 or 107 psi.

To counter the radial stress is the tensile strength of the basalt rock ceiling, the weight of the ceiling blocks and the weight of regolith that will cover the house.  Although basalt can vary greatly in its characteristics, a typical tensile strength for basalt has been given as -14.5 +/- 3.3 MPa, (3) which converts to 2103 psi.   Based only on this consideration, the safety factor would be 14,500,000 / 2103 or 6895 to 1.   The weight of the stone block and the weight of regolith would be negligible.  However, this is not the only consideration; we know that the cracks between stones or any other cracks would allow air to escape very quickly, as we can experience if we remove the valve stem from a tire.  Air containment will depend entirely upon keeping the cracks small and applying epoxy resin to cracks when they are discovered.

 Equations

Return to the article Pioneer House Building.

 

References

  1.  “Engineering Toolbox,” http://www.engineeringtoolbox.com
  2. “Thick Walled Cylinders,” http://courses.washington.edu/me354a/Thick%20Walled%20Cylinders.pdf
  3. (3 R.A. Schultz, “Limits of strength and deformation properties of jointed basaltic masses,” Rock Mechanics and Rock Engineering (1995) 28 (1), 1-15.

 (4) “Densities of common rock types,” http://geology.about.com/cs/rock_types/a/aarockspecgrav.htm